Identify the oxidizing agent and medium

In alkaline medium, permanganate ion $\mathrm{MnO_4^-}$ acts as a strong oxidizing agent.

In basic medium, $\mathrm{MnO_4^-}$ is typically reduced to $\mathrm{MnO_2}$:

$\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$

Possible oxidation of iodide

Iodide ion $\mathrm{I^-}$ can be oxidized to different products depending on the oxidizing conditions.

In alkaline medium with a strong oxidizing agent like permanganate, iodide is oxidized beyond $I_2$ and forms iodate, $\mathrm{IO_3^-}$.

Verify by half-reaction method

Oxidation half-reaction of iodide to iodate in basic medium:

Start in acidic form first:

$\mathrm{I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-}$

Convert to basic medium by adding $6OH^-$ to both sides:

$\mathrm{I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-}$

Combine with permanganate reduction half-reaction

Reduction half-reaction:

$\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$

Multiply this by $2$ to balance electrons:

$\mathrm{2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-}$

Add with iodide oxidation:

$\mathrm{I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-}$

On adding and canceling common terms:

$\mathrm{I^- + 2MnO_4^- + H_2O \rightarrow IO_3^- + 2MnO_2 + 2OH^-}$

This confirms that $\mathrm{I^-}$ is oxidized to $\mathrm{IO_3^-}$.

Evaluate options

• A: $I_2$ — not the final product in alkaline medium with permanganate.
• B: $\mathrm{IO_3^-}$ — correct.
• C: $\mathrm{IO^-}$ — not the usual oxidation product here.
• D: $\mathrm{IO_4^-}$ — oxidation too far for this reaction under standard alkaline conditions.

Final answer

$\boxed{\mathrm{IO_3^-}}$

So, the correct option is B.